Draw the graph of $| x | + | y | + | x + y | = 1$ .

Open in App

Solution

We know that,
$| x | = {\begin{matrix} x & x \geq 0 - x & x < 0 \end{matrix}$
$| y | = {\begin{matrix} y & y \geq 0 - y & y < 0 \end{matrix}$
$| x + y | = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - x - y & x < 0, & y < 0 - x + y & x < 0, & y \geq 0 x - y & x \geq 0, & y < 0 x + y & x \geq 0, & y \geq 0 \end{matrix}$
So, we can write $| x | + | y | + | x + y | = 1$
as $⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + y + x + y = 1 - x + y - x + y = 1 x - y + x - y = 1 - x - y - x - y = 1 \end{matrix} = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + y = \frac{1}{2} y - x = \frac{1}{2} x - y = \frac{1}{2} x + y = - \frac{1}{2} \end{matrix}$ .

Suggest Corrections

Similar questions

\frac{y}{x} = \frac{y + 1}{x + 2}

y = | x |^{\frac{1}{2}}

- 1 \leq x \leq 1.

(1) Draw the graph of y = x² and y = 2x + 3 and hence solve the equation x² − 2x − 3 = 0.

(2) Draw the graph of y = 2x²and y = 3 − x and hence solve the equation 2x² + x − 3 = 0.

(3) Draw the graph of y = 2x² and y = 3 + x and hence solve the equation 2x² − x − 3 = 0.

x - y = 0

x = - 1

y = - - -

Join BYJU'S Learning Program